Holt Algebra 1 6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
Holt Algebra 1
6-2 Solving Systems by Substitution 6-2 Solving Systems by Substitution
Holt Algebra 1
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 1
6-2 Solving Systems by Substitution
Warm Up Solve each equation for x.
1. y = x + 3
2. y = 3x – 4
Simplify each expression.
x = y – 3
2x – 10 3. 2(x – 5)
4. 12 – 3(x + 1) 9 – 3x
Holt Algebra 1
6-2 Solving Systems by Substitution
Warm Up Continued Evaluate each expression for the given value of x.
5. x + 8 for x = 6
6. 3(x – 7) for x =10
12
9
Holt Algebra 1
6-2 Solving Systems by Substitution
Solve linear equations in two variables by substitution.
Objective
Holt Algebra 1
6-2 Solving Systems by Substitution
Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution.
The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.
Holt Algebra 1
6-2 Solving Systems by Substitution
Solving Systems of Equations by Substitution
Step 2
Step 3
Step 4
Step 5
Step 1 Solve for one variable in at least one equation, if necessary.
Substitute the resulting expression into the other equation.
Solve that equation to get the value of the first variable.
Substitute that value into one of the original equations and solve.
Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.
Holt Algebra 1
6-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1A: Solving a System of Linear Equations by
Substitution
y = 3x
y = x – 2
Step 1 y = 3x
y = x – 2
Both equations are solved for y.
Step 2 y = x – 2
3x = x – 2
Substitute 3x for y in the second
equation.
Solve for x. Subtract x from both
sides and then divide by 2.
Step 3 –x –x
2x = –2 2x = –2
2 2 x = –1
Holt Algebra 1
6-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1A Continued
Step 4 y = 3x Write one of the original
equations. Substitute –1 for x. y = 3(–1)
y = –3
Step 5 (–1, –3)
Check Substitute (–1, –3) into both equations in the system.
Write the solution as an ordered pair.
y = 3x
–3 3(–1)
–3 –3
y = x – 2
–3 –1 – 2
–3 –3
Holt Algebra 1
6-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1B: Solving a System of Linear Equations by
Substitution
y = x + 1
4x + y = 6
Step 1 y = x + 1 The first equation is solved for y.
Step 2 4x + y = 6
4x + (x + 1) = 6 Substitute x + 1 for y in the
second equation.
Subtract 1 from both sides. Step 3 –1 –1
5x = 5
5 5 x = 1
5x = 5
5x + 1 = 6 Simplify. Solve for x.
Divide both sides by 5.
Holt Algebra 1
6-2 Solving Systems by Substitution
Solve the system by substitution.
Example1B Continued
Step 4 y = x + 1 Write one of the original
equations. Substitute 1 for x. y = 1 + 1
y = 2
Step 5 (1, 2)
Check Substitute (1, 2) into both equations in the system.
Write the solution as an ordered pair.
y = x + 1
2 1 + 1
2 2
4x + y = 6
4(1) + 2 6
6 6
Holt Algebra 1
6-2 Solving Systems by Substitution
Solve the system by substitution.
Example 1C: Solving a System of Linear Equations by
Substitution
x + 2y = –1
x – y = 5
Step 1 x + 2y = –1 Solve the first equation for x by
subtracting 2y from both sides.
Step 2 x – y = 5
(–2y – 1) – y = 5
Substitute –2y – 1 for x in the
second equation.
–3y – 1 = 5 Simplify.
−2y −2y
x = –2y – 1
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 1C Continued
Step 3 –3y – 1 = 5 Add 1 to both sides. +1 +1
–3y = 6
–3y = 6
–3 –3
y = –2
Solve for y.
Divide both sides by –3.
Step 4 x – y = 5
x – (–2) = 5
x + 2 = 5 –2 –2
x = 3
Step 5 (3, –2)
Write one of the original equations.
Substitute –2 for y.
Subtract 2 from both sides.
Write the solution as an ordered pair.
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 1a
Solve the system by substitution.
y = x + 3
y = 2x + 5
Both equations are solved for y. Step 1 y = x + 3
y = 2x + 5
Substitute 2x + 5 for y in the first
equation.
Solve for x. Subtract x and 5
from both sides. –x – 5 –x – 5
x = –2
Step 3 2x + 5 = x + 3
Step 2
2x + 5 = x + 3
y = x + 3
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 1a Continued
Solve the system by substitution.
Step 4 y = x + 3 Write one of the original
equations. Substitute –2 for x. y = –2 + 3
y = 1
Step 5 (–2, 1) Write the solution as an ordered pair.
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 1b
Solve the system by substitution.
x = 2y – 4
x + 8y = 16
The first equation is solved for x. Step 1 x = 2y – 4
Substitute 2y – 4 for x in the
second equation.
Simplify. Then solve for y.
(2y – 4) + 8y = 16
x + 8y = 16 Step 2
Step 3 10y – 4 = 16
Add 4 to both sides. +4 +4 10y = 20
y = 2
10y 20
10 10 = Divide both sides by 10.
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 1b Continued
Solve the system by substitution.
Step 4 x + 8y = 16 Write one of the original equations.
Substitute 2 for y. x + 8(2) = 16
x + 16 = 16
x = 0
– 16 –16 Simplify.
Subtract 16 from both sides.
Step 5 (0, 2) Write the solution as
an ordered pair.
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 1c
Solve the system by substitution.
2x + y = –4
x + y = –7
Solve the second equation for x
by subtracting y from each
side.
Substitute –y – 7 for x in the
first equation.
Distribute 2.
2(–y – 7) + y = –4
x = –y – 7 Step 2
Step 1 x + y = –7 – y – y
x = –y – 7
2(–y – 7) + y = –4
–2y – 14 + y = –4
Holt Algebra 1
6-2 Solving Systems by Substitution
Combine like terms. Step 3
+14 +14
–y = 10
Check It Out! Example 1c Continued
Solve the system by substitution.
–2y – 14 + y = –4
Add 14 to each side.
–y – 14 = –4
y = –10
Step 4 x + y = –7 Write one of the original equations.
Substitute –10 for y. x + (–10) = –7
x – 10 = – 7
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 1c Continued
Solve the system by substitution.
x – 10 = –7 Step 5
+10 +10
x = 3
Add 10 to both sides.
Step 6 (3, –10) Write the solution as an
ordered pair.
Holt Algebra 1
6-2 Solving Systems by Substitution
Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.
Holt Algebra 1
6-2 Solving Systems by Substitution
When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved.
Caution
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2: Using the Distributive Property
y + 6x = 11
3x + 2y = –5 Solve by substitution.
Solve the first equation for y
by subtracting 6x from each
side.
Step 1 y + 6x = 11 – 6x – 6x
y = –6x + 11
Substitute –6x + 11 for y in the
second equation.
Distribute 2 to the expression
in parenthesis.
3x + 2(–6x + 11) = –5
3x + 2y = –5 Step 2
3x + 2(–6x + 11) = –5
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 3
Example 2 Continued
y + 6x = 11
3x + 2y = –5 Solve by substitution.
3x + 2(–6x) + 2(11) = –5
–9x + 22 = –5
Simplify. Solve for x.
Subtract 22 from
both sides. –9x = –27
– 22 –22
Divide both sides
by –9.
–9x = –27
–9 –9
x = 3
3x – 12x + 22 = –5
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 4 y + 6x = 11
Substitute 3 for x. y + 6(3) = 11
Subtract 18 from each side. y + 18 = 11
–18 –18
y = –7
Step 5 (3, –7) Write the solution as an
ordered pair.
Simplify.
Example 2 Continued
y + 6x = 11
3x + 2y = –5 Solve by substitution.
Write one of the original
equations.
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 2
–2x + y = 8
3x + 2y = 9 Solve by substitution.
Solve the first equation for y
by adding 2x to each side.
Step 1 –2x + y = 8 + 2x +2x
y = 2x + 8
Substitute 2x + 8 for y in the
second equation. 3x + 2(2x + 8) = 9
3x + 2y = 9 Step 2
Distribute 2 to the expression
in parenthesis.
3x + 2(2x + 8) = 9
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 3 3x + 2(2x) + 2(8) = 9
7x + 16 = 9
Simplify. Solve for x.
Subtract 16 from
both sides. 7x = –7
–16 –16
Divide both sides
by 7.
7x = –7
7 7
x = –1
Check It Out! Example 2 Continued
–2x + y = 8
3x + 2y = 9 Solve by substitution.
3x + 4x + 16 = 9
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 4 –2x + y = 8
Substitute –1 for x. –2(–1) + y = 8
y + 2 = 8
–2 –2
y = 6
Step 5 (–1, 6) Write the solution as an
ordered pair.
Check It Out! Example 2 Continued
–2x + y = 8
3x + 2y = 9 Solve by substitution.
Subtract 2 from each side.
Simplify.
Write one of the original
equations.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2: Consumer Economics Application
Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Write an equation for each option. Let t represent the total amount paid and m represent the number of months.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2 Continued
Total paid is
sign-up fee plus
payment amount
for each month.
Option 1 t = $50 + $20 m
Option 2 t = $30 + $25 m
Step 1 t = 50 + 20m
t = 30 + 25m
Both equations are solved
for t.
Step 2 50 + 20m = 30 + 25m Substitute 50 + 20m for t in
the second equation.
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 3 50 + 20m = 30 + 25m Solve for m. Subtract 20m
from both sides. –20m – 20m
50 = 30 + 5m Subtract 30 from both
sides. –30 –30
20 = 5m Divide both sides by 5.
Write one of the original
equations.
Step 4 t = 30 + 25m
t = 30 + 25(4)
t = 30 + 100
t = 130
Substitute 4 for m.
Simplify.
Example 2 Continued
5 5
m = 4
20 = 5m
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 5 (4, 130) Write the solution as an
ordered pair.
In 4 months, the total cost for each option would be the same $130.
Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.
Example 2 Continued
Option 1: t = 50 + 20(12) = 290
Option 2: t = 30 + 25(12) = 330
If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.
Holt Algebra 1
6-2 Solving Systems by Substitution
Check It Out! Example 3
One cable television provider has a $60 setup fee and $80 per month, and the second has a $160 equipment fee and $70 per month.
a. In how many months will the cost be the same? What will that cost be.
Write an equation for each option. Let t represent the total amount paid and m represent the number of months.
Holt Algebra 1
6-2 Solving Systems by Substitution
Total paid is fee plus
payment amount
for each month.
Option 1 t = $60 + $80 m
Option 2 t = $160 + $70 m
Check It Out! Example 3 Continued
Step 1 t = 60 + 80m
t = 160 + 70m
Both equations are solved
for t.
Step 2 60 + 80m = 160 + 70m Substitute 60 + 80m for t in
the second equation.
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 3 60 + 80m = 160 + 70m Solve for m. Subtract 70m
from both sides. –70m –70m
60 + 10m = 160 Subtract 60 from both
sides.
Divide both sides by 10.
–60 –60
10m = 100
10 10 m = 10
Write one of the original
equations.
Step 4 t = 160 + 70m
t = 160 + 70(10)
t = 160 + 700
t = 860
Substitute 10 for m.
Simplify.
Check It Out! Example 3 Continued
Holt Algebra 1
6-2 Solving Systems by Substitution
Step 5 (10, 860) Write the solution as an ordered pair.
In 10 months, the total cost for each option would be the same, $860.
The first option is cheaper for the first six months.
Check It Out! Example 3 Continued
Option 1: t = 60 + 80(6) = 540
Option 2: t = 160 + 270(6) = 580
b. If you plan to move in 6 months, which is the cheaper option? Explain.
Holt Algebra 1
6-2 Solving Systems by Substitution
Lesson Quiz: Part I
Solve each system by substitution.
1.
2.
3.
(1, 2)
(–2, –4) y = 2x
x = 6y – 11
3x – 2y = –1
–3x + y = –1
x – y = 4
Holt Algebra 1
6-2 Solving Systems by Substitution
Lesson Quiz: Part II
4. Plumber A charges $60 an hour. Plumber B charges $40 to visit your home plus $55 for each hour. For how many hours will the total cost for each plumber be the same? How much will that cost be? If a customer thinks they will need a plumber for 5 hours, which plumber should the customer hire? Explain.
8 hours; $480; plumber A: plumber A is cheaper for less than 8 hours.